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$ (sec\theta - tan\theta )^2 ( 1 + sin\theta)^2 ÷ cos^2 \theta $ = ? |
$cos^2\theta $ 1 $cot^2\theta $ -1 |
1 |
Put θ = 30º Now , (secθ - tanθ)2 (1 + sinθ)2 ÷ cos2θ = (√2 - 1)2 (1 + \(\frac{1}{√2}\) )2 ÷ 2 = (√2 - 1)2 (√2 +1)2 = ( 2 + 1 - 2√2) ( 2 + 1 + 2√2) = (3- 2√2) ( 3 + 2√2) = 9 - 8 = 1 θ |
